Locks and keys

1. Step 1 · Every pair must be stopped by some lock

   $\displaystyle a \text{ lock neither of the } 2 \text{ has } a \text{ key to}$

   No 2 partners can open the safe, so for each pair there’s a lock that both of them are missing.

2. Step 2 · That lock’s keys go to everyone else

   $\displaystyle \text{all } 3 \text{ of the others}$

   If any of the other 3 also lacked that key, a trio could be stuck. So the lock stops exactly one pair and no other.

3. Step 3 · One lock for each pair

   $\displaystyle \binom{5}{2} = 5 \times 4 \div 2 = 10$

   Each pair needs its own lock, and a lock per pair is enough: any trio includes someone outside every pair, so it can open all of them.

Answer$\displaystyle 10$